1) The purpose of the lecture is to teach how to index and determine cell parameters from SAED patterns, determine possible space groups and point groups, and solve simple structures from PED patterns. 2) Example materials used are aluminum and rutile-type SnO2. The slides and exercises for determining unknown cell parameters from SAED patterns are provided. 3) The document provides a step-by-step example of determining the cell parameters of aluminum from its SAED patterns, starting with no prior knowledge of the material. The correct cell is determined to be cubic with a=b=c=4.06 Angstroms.

2. Purpose of this lecture
At the end of this lecture, you should be able to
1) index SAED patterns if the cell parameters are known
2) know how to determine unknown cell parameters
from SAED patterns
3) determine the possible space groups from SAED
patterns
4) determine possible point groups from CBED patterns
5) solve a simple structure ab initio from PED patterns

3. Example materials used in this lecture
Aluminum
Rutile-type SnO2
Slides are on http://www.slideshare.net/johader
The pages with the ED for the exercises can be
found at the end of this ppt.

4. 1. Selected area
electron diffraction
(SAED)

5. Example 1: a known material, e.g. SnO2
Tetragonal, a=b= 4.72 Å, c=3.16 Å
4.72
3.34 2.63
4.72 2.36
1.58

6. List of hkl – d
Powdercell
Input:
P42/mnm
a=b=4.72 Å
c= 3.17 Å

7. Determine the zone-axis
010
110 101
100 200
002
[001] [010]

8. If cell parameters/material unknown
Shortest vectors, high symmetry, ...
best during TEM experiment

9. Starting from nothing: aluminum
First determine the cell parameters
Make a tilt series

10. Do you recognize any typical symmetry?
Only 2-fold
2- and 4-fold
2-, 4- and 6-fold

11. Do you recognize any typical symmetry?
Only 2-fold
2- and 4-fold
2-, 4- and 6-fold

12. What crystal class has both 4-fold and 6-fold axes?
cubic
hexagonal
none

13. What crystal class has both 4-fold and 6-fold axes?
cubic
hexagonal
none

14. In SAED patterns there is always an
inversion centre due to the diffraction
geometry!
+ =

15. What crystal class has both 4-fold and 3-fold axes?
cubic
hexagonal
none

16. What crystal class has both 4-fold and 3-fold axes?
cubic
hexagonal
none

17. b
a
c

18. How many possible choices for the basic vectors are
there on this square pattern?
1
2
more

19. How many possible choices for the basic vectors are
there on this square pattern?
1
2
more

20. There can be a reflection condition!
hk0: 100 200
a*
020 a* h=2n,
no cond. 010
b* b* k=2n
020 200
b* a* Choose simplest
h+k=2n first and see if ok
with other zones.

21. hk0: 100
a*
no cond. 010
b*
Measure R
Calculate d from R.d=λL (known)
Determine a, b, c
Make list of d-spacings
Check if you can index the other patterns with this

22. What will be the value of a and b with this choice?
1.43
2.03
4.06
0.78
2.03 1.43 1.22
a*
2.03 2.03 2.03
b*
0.91
2.34 2.34
2.03 2.03
1.43

23. What will be the value of a and b with this choice?
1.43
2.03
4.06
0.78
2.03 1.43 1.22
a*
2.03 2.03 2.03
b*
0.91
2.34 2.34
2.03 2.03
1.43

24. If no reflection condition:
a=b= c=2.03 Å
list of d-spacings freeware: Powdercell
H K L d/Å
_____________ Can you index all
observed
1 0 0 2.03000 reflections using
1 1 0 1.43543 these cell
1 1 1 1.17202 parameters?
2 0 0 1.01500
yes
2 1 0 0.90784
2 1 1 0.82874 no

25. If no reflection condition:
a=b= c=2.03 Å
list of d-spacings freeware: Powdercell
H K L d/Å
_____________ Can you index all
observed
1 0 0 2.03000 reflections using
1 1 0 1.43543 these cell
1 1 1 1.17202 parameters?
2 0 0 1.01500
yes
2 1 0 0.90784
2 1 1 0.82874 no

26. hk0: 100 200
a*
020 a* h=2n,
no cond. 010
b* b* k=2n
Increase the cell
020 200 parameters to the next
b* a* possibility...
h+k=2n
Which one has second
smallest cell parameters?
top
left

27. hk0: 100 200
a*
020 a* h=2n,
no cond. 010
b* b* k=2n
Increase the cell
020 200 parameters to the next
b* a* possibility...
h+k=2n
Which one has second
smallest cell parameters?
top
left

28. 020 2.03 200
1.43 1.43
b* a*
h+k=2n
2.03
In this case, a=b=
1.43 Å
2.03 Å
2.86 Å

29. 020 2.03 200
1.43 1.43
b* a*
h+k=2n
2.03
In this case, a=b=
1.43 Å
2.03 Å
2.86 Å

30. If hk0:h+k=2n, then a=b= 2.86 Å
list of d-spacings, Powdercell:
H K L d/Å
______________
1 0 0 2,86000
1 1 0 2,02233 Can you index all
1 1 1 1,65122 observed
2 0 0 1,43000
2 1 0 1,27903 reflections using
2 1 1 1,16759
2 2 0 1,01116
these cell
3 0 0 0,95333 parameters?
2 2 1 0,95333
3 1 0 0,90441 yes
3 1 1 0,86232 no
2 2 2 0,82561
3 2 0 0,79322

31. If hk0:h+k=2n, then a=b= 2.86 Å
list of d-spacings, Powdercell:
H K L d/Å
______________
1 0 0 2,86000
1 1 0 2,02233 Can you index all
1 1 1 1,65122 observed
2 0 0 1,43000
2 1 0 1,27903 reflections using
2 1 1 1,16759
2 2 0 1,01116
these cell
3 0 0 0,95333 parameters?
2 2 1 0,95333
3 1 0 0,90441 yes
3 1 1 0,86232 no
2 2 2 0,82561
3 2 0 0,79322

32. If hk0:h+k=2n, then a=b= 2.86 Å
list of d-spacings, Powdercell:
H K L d/Å I show also the
______________
1 0 0 2,86000 reflections not in
1 1 0 2,02233
1 1 1 1,65122
agreement with
2 0 0 1,43000 hk0: h+k=2n.
2 1 0 1,27903
2 1 1 1,16759 Will they be seen on any
2 2 0 1,01116 of the SAED patterns of
3 0 0 0,95333
2 2 1 0,95333 the tilt series?
3 1 0 0,90441
3 1 1 0,86232
it is possible
2 2 2 0,82561 always
3 2 0 0,79322
never

33. If hk0:h+k=2n, then a=b= 2.86 Å
list of d-spacings, Powdercell:
H K L d/Å I show also the
______________
1 0 0 2,86000 reflections not in
1 1 0 2,02233
1 1 1 1,65122
agreement with
2 0 0 1,43000 hk0: h+k=2n.
2 1 0 1,27903
2 1 1 1,16759 Will they be seen on any
2 2 0 1,01116 of the SAED patterns of
3 0 0 0,95333
2 2 1 0,95333 the tilt series?
3 1 0 0,90441
3 1 1 0,86232
it is possible
2 2 2 0,82561 always
3 2 0 0,79322
never

34. hk0: 100 200
a*
020 a* h=2n,
no cond. 010
b* b* k=2n
020 200
Next possibility...
b* a*
h+k=2n

35. 200
2.03
020 a*
2.03
b*
In this case, a=b=
2.03 Å
4.06 Å
5.92 Å

36. 200
2.03
020 a*
2.03
b*
In this case, a=b=
2.03 Å
4.06 Å
5.92 Å

37. If hk0:h,k=2n, then a=b= 4.06 Å
list of d-spacings, Powdercell:
H K L d/Å 3 2 0 1,12604 4 3 1 0,79623
___________ 3 2 1 1,08508 5 1 1 0,78135
4 0 0 1,01500 3 3 3 0,78135
1 0 0 4,06000 4 1 0 0,98469
1 1 0 2,87085 3 2 2 0,98469
1 1 1 2,34404 4 1 1 0,95695 Can you index all
2 0 0 2,03000 3 3 0 0,95695 observed reflections
2 1 0 1,81569 3 3 1 0,93143 with these cell
2 1 1 1,65749 4 2 0 0,90784 parameters?
2 2 0 1,43543 4 2 1 0,88596
3 0 0 1,35333 3 3 2 0,86559
yes
2 2 1 1,35333 4 2 2 0,82874
3 1 0 1,28388 5 0 0 0,81200 no
3 1 1 1,22414 4 3 0 0,81200
2 2 2 1,17202 5 1 0 0,79623 Do it now.

38. If hk0:h,k=2n, then a=b= 4.06 Å
list of d-spacings, Powdercell:
H K L d/Å 3 2 0 1,12604 4 3 1 0,79623
___________ 3 2 1 1,08508 5 1 1 0,78135
4 0 0 1,01500 3 3 3 0,78135
1 0 0 4,06000 4 1 0 0,98469
1 1 0 2,87085 3 2 2 0,98469
1 1 1 2,34404 4 1 1 0,95695 Can you index all
2 0 0 2,03000 3 3 0 0,95695 observed reflections
2 1 0 1,81569 3 3 1 0,93143 with these cell
2 1 1 1,65749 4 2 0 0,90784 parameters?
2 2 0 1,43543 4 2 1 0,88596
3 0 0 1,35333 3 3 2 0,86559
yes
2 2 1 1,35333 4 2 2 0,82874
3 1 0 1,28388 5 0 0 0,81200 no
3 1 1 1,22414 4 3 0 0,81200
2 2 2 1,17202 5 1 0 0,79623 Do it now.

39. 110
The index of this reflection is 220
440
0.78
1.43 1.22
200
020 2.03 2.03
0.91
2.34 2.34
2.03 2.03
1.43

40. 110
The index of this reflection is 220
440
0.78
1.43 1.22
200
020 2.03 2.03
0.91
2.34 2.34
2.03 2.03
1.43

41. 200
The index of this reflection is 010
020
220 200 0.78
1.22
020 2.03 2.03
0.91
2.34 2.34
2.03 2.03
1.43

42. 200
The index of this reflection is 010
020
220 200 0.78
1.22
020 2.03 2.03
0.91
2.34 2.34
2.03 2.03
1.43

43. 333 How to
The index of this reflection is 511 decide
-
511 this?
220 200 0.78
1.22
020 020 020
0.91
2.34 2.34
020 020
1.43

44. 333 How to
The index of this reflection is 511 decide
-
511 this?
220 200 0.78
1.22
020 020 020
0.91
2.34 2.34
020 020
1.43

45. 333
The index of this reflection is 511
-
511
020

46. - -
511 and 511? 115 and 115?
- -- - --
511 and 511? 115 and 115?
Calculate angles etc (lectures Mon-Wed)
or
just do it graphically...

47. b* a*
100 200 300 400 500
c*
001 101 201 301 401 501
002 102 202 302 402 502
003 103 203 303 403 503

48. 333
The index of this reflection is 511
-
511
-
511 511
020

49. 511 -
220 200 511
1.22
020 020 020
0.91
2.34 2.34
020 020
1.43

50. b* a*
100 200 300 400 500
c*
001 101 201 301 401 501
002 102 202 302 402 502
003 103 203 303 403 503

51. 411
The index of this reflection is 311
511 - 401
220 200 511
1.22
020 020 020
0.91
2.34 2.34
020 020
1.43

52. 411
The index of this reflection is 311
511 - 401
220 200 511
1.22
020 020 020
0.91
2.34 2.34
020 020
1.43

53. 511 -
220 200 511
311
020 020 020
0.91
2.34 2.34
020 020
1.43

54. b* a*
100 200 300 400 500
c*
001 101 201 301 401 501
002 102 202 302 402 502
003 103 203 303 403 503

55. 511 -
220 200 511
311
020 020 020
402
2.34 2.34
020 020
1.43

56. 511 -
220 200 511
311
020 020 020
402
111 -
111
020 020

57. 511 -
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

58. Zone axes for this zone for example? 103
301
-
511 103
-
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

59. Zone axes for this zone for example? 103
301
-
511 103
-
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

60. -
[105]
[001] -
[103]
511 -
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202
- -
[101]
[102]

61. Indexed Space group?
-> cell parameter: Possibility from SAED:
a=4.06 Å extinction symbol.
-> Determine reflection
conditions.

62. hkl: h+k+l=2n Bravais lattice: F
all odd, all even P
no conditions I
511 -
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

63. hkl: h+k+l=2n Bravais lattice: F
all odd, all even P
no conditions I
511 -
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

64. hkl: h+k+l=2n Bravais lattice: F
all odd, all even P
no conditions I
511 -
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

65. We know: hk0: h=2n, k=2n. Is this an extra
condition, or is it a consequence of F also?
Consequence
511 - Extra
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

66. We know: hk0: h=2n, k=2n. Is this an extra
condition, or is it a consequence of F also?
Consequence
511 - Extra
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

67. Are there àny extra conditions, i.e. conditions
that are not already implied by hkl: all odd, all
even? Yes
No 511 -
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

68. Are there àny extra conditions, i.e. conditions
that are not already implied by hkl: all odd, all
even? Yes
No 511 -
220 200 511
311
020 020 020
402
111 -
111
020 020
- -
202

69. - -
- F432, F43m, Fm3m
Possible space groups: F23, Fm3,
from: International Tables for Crystallography Volume A, ed. Theo Hahn, Kluwer Ac. Pub., 2002

70. - -
- F432, F43m, Fm3m
Possible space groups: F23, Fm3,
Same extinctions.
- 2, m; give no extinctions.
Difference: 1,
Convergent beam electron diffraction
CBED

71. 2. Convergent beam
electron diffraction
(CBED)

72. SAED CBED
Example: rutile-type SnO2

73. Projection symmetry: 2D, diffuse features
Full symmetry: 3D, sharp features
BF
WP
Lots of tables!

75. 6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
21R
2R
2
1R
1
4/mmm
6/mmm
mmm
-42m
-43m
-6m2
4mm
6mm
m3m
mm2
2/m
4/m
6/m
-3m
222
422
622
432
3m
m3
32
23
-1
-4
-3
-6
m
1
2
4
3
6

76. WP
proj
WP

77. Projection whole pattern symmetry [001]
4
4mm
2mm

78. Projection whole pattern symmetry [001]
4
4mm
2mm

79. Projection WP: 4mm
Projection diffraction group:
Table Eades
4
4RmmR
4mm1R
Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992

80. Projection WP: 4mm
Projection diffraction group:
Table Eades
4
4RmmR
4mm1R
Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992

81. Projection diffraction group:
4mm1R
Possible diffraction groups:
4mRmR
4mm
4RmmR
4mm1R
Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992

82. 6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
Table redrawn 21R
from B.F. Buxton, 2R
J.A. Eades, J.W. 2
Steeds, G.M. 1R
Rackham: Phil.
1
Trans. R. Soc.
4/mmm
6/mmm
London, 281
mmm
-42m
-43m
-6m2
4mm
6mm
m3m
mm2
2/m
4/m
6/m
-3m
222
422
622
432
(1976) 171
3m
m3
32
23
-1
-4
-3
-6
m
1
2
4
3
6

83. [...] [...]
What will be useful to narrow it down further?
look at the bright field symmetry
look at the whole pattern symmetry

84. [...] [...]
What will be useful to narrow it down further?
look at the bright field symmetry
look at the whole pattern symmetry

85. WP symmetry
4
4mm
2mm

86. WP symmetry
4
4mm
2mm

87. [...] [...]

88. 6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
21R
2R
2
1R
1
4/mmm
6/mmm
mmm
-42m
-43m
-6m2
4mm
6mm
m3m
mm2
2/m
4/m
6/m
-3m
222
422
622
432
3m
m3
32
23
-1
-4
-3
-6
m
1
2
4
3
6

89. Another pattern
Index it.
4.72 =010
This is: 2.63
-
=101
[101]
[110]
[111]
SAED

90. Another pattern
Index it.
4.72 =010
This is: 2.63
-
=101
[101]
[110]
[111]
SAED

91. Projection whole pattern [101]
(smaller cond.ap.)
2
m
2mm

92. Projection whole pattern [101]
(smaller cond.ap.)
2
m
2mm

93. Possible projection
diffraction group:
21R
m1R
2mm1R

94. Possible projection
diffraction group:
21R
m1R
2mm1R

95. Whole pattern [101]
(smaller cond.ap.)
2
m
2mm

96. Whole pattern [101]
(smaller cond.ap.)
2
m
2mm

97. Diffraction group:
2mm
2RmmR
2mm1R

98. Diffraction group:
2mm
2RmmR
2mm1R

99. 6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
21R
2R
2
1R
1
4/mmm
6/mmm
mmm
-42m
-43m
-6m2
4mm
6mm
m3m
mm2
2/m
4/m
6/m
-3m
222
422
622
432
3m
m3
32
23
-1
-4
-3
-6
m
1
2
4
3
6

100. Possible point groups
4/mmm
m3m

101. CBED
from: International Tables for Crystallography Volume A, ed. Theo Hahn, Kluwer Ac. Pub., 2002

102. Combine with information about
reflection conditions from SAED patterns
020 002 002
110 011
200 020 -
110
hkl:
no conditions
h+k+l=2n

103. Combine with information about
reflection conditions from SAED patterns
020 002 002
110 011
200 020 -
110
hkl:
no conditions
h+k+l=2n

104. Combine with information about
reflection conditions from SAED patterns
020 002 002
110 011
200 020 -
110
hk0: 0kl: hhl:
no conditions no conditions no conditions
h+k=2n k+l=2n l=2n
k=2n or l=2n

105. Combine with information about
reflection conditions from SAED patterns
020 002 002
110 011
200 020 -
110
hk0: 0kl: hhl:
no conditions no conditions no conditions
h+k=2n k+l=2n l=2n
k=2n or l=2n

106. Combine with information about
reflection conditions from SAED patterns
020 002 002
110 011
200 020 -
110
hk0: 0kl: hhl:
no conditions no conditions no conditions
h+k=2n k+l=2n l=2n
k=2n or l=2n

107. Combine with information about
reflection conditions from SAED patterns
020 002 002
110 011
200 020 -
110
hk0: 0kl: hhl:
no conditions no conditions no conditions
h+k=2n k+l=2n l=2n
k=2n or l=2n

108. International Tables
CBED
SAED
Space Group P42/mnm

109. 3. Precession
electron diffraction
(PED)

110. Ab initio solution of structures using electron
diffraction:
You have as
experimental data: You get:
only electron the structure
diffraction patterns
Trouble: dynamical diffraction
110

111. Precession electron diffraction
Example: SnO2
Demonstration of the precession technique with EMAP

114. Intensities of the reflections
from several zones
+
cell parameters and space
STRUCTURE
group from SAED-CBED
+
composition from EDX/EELS
or nominal composition

115. Extract the intensities using
EDM
EXTRAX
CRISP
...
Possibilities to introduce
mistakes:
not well oriented
too thick
too small precession angle
too large precession angle
…

116. Apply symmetry

117. Combine separate lists
117 Many possibilities to introduce mistakes!!

118. Direct methods, or optimisation methods
made for single crystal data
SIR2008
Fox
Endeavour
...
Right structure càn come out (GIGO).

119. Purpose of this lecture
At the end of this lecture, you should be able to
1) index SAED patterns if the cell parameters are known
2) know how to determine unknown cell parameters
from SAED patterns
3) determine the possible space groups from SAED
patterns
4) determine possible point groups from CBED patterns
5) determine a simple structure ab initio from PED
patterns

121. ED patterns for the exercises
These ED are the same as those on the
previous slides, but the contrast has
been reversed to enable easy writing.

125. SAED and CBED SnO2 second zone